I need help proving a simple premise regarding primes. I am not a mathematician, just a math lover. I found a very simple function that seems to generate prime numbers, and numbers that are the product of primes beginning with $7$ and $17$.
I found this simple function while looking for a prime number generator. I am not suggesting any computer use for this function. I just want to know if it could be proven false?
For any non zero integer number $x$, $F(x)=2x^2-1$ . $F(x)$ is always a prime number or a number that is the product of primes that are $7$ or larger.
I really do not know if this is important to others. It is to me, because it could explain some prime numbers patterns within nature. The squaring could be seen as accretion, and the subtraction is quanta of radiation. This occurs within a particular space that has the geometry of tightly packed spheres that are randomly spinning and rotating around each other, representing the vacuum of space.
This is the best that I can do. I hope it is sufficient. Sorry for the previous post.
 A: To say a positive integer is "a prime or product of primes that are $7$ or larger" isn't a very strong statement: it simply means that the integer is not divisible by $2$, $3$ or $5$ (unless it equals one of them).
It's easy to see $F(x)$ is not divisible by $2$ (i.e. is odd), since it is $1$ less than $2x^2$, and $2x^2$ is even.
To show the other two statements, you need to know some basic techniques of modular arithmetic. This is essentially thinking about a number in terms of its remainder when divided by some specific value. Take division by $3$ as an example. When we divide $x$ by $3$ we get a remainder of $0$, $1$ or $2$. This means $x$ can be written as one of the following:

*

*$x=3y$,

*$x=3y+1$, or

*$x=3y+2$.

In the first case, $F(x)=2\times(3y)^2-1=18y^2-1$ is not a multiple of $3$ (it is $1$ less than a multiple of $3$). In the second, $F(x)=2\times(3y+1)^2-1=18y^2+12y+1$, which is $1$ more than a multiple of $3$. The third case gives $18y^2+24y+7$, which is $1$ more than a multiple of $3$.
To show it's not divisible by $5$ you can do a similar thing with remainders after dividing by $5$, which has $5$ cases.
A: You would like to prove that $ p \not | 2x^2 - 1, \ p=3, 5.$
Assume otherwise that  $ 3  \  | 2x^2 - 1$ for some $x$. In the language of congruences, we write $2x^2 \equiv 1 \ $(mod $\ 3$) ,  $2\cdot 2 x^2 \equiv 2 \cdot 1 \ $(mod$  \ 3) $,
$x^2 \equiv 2 \ $ (mod $3) $ which is not possible because every perfect square is congruent to 0 or 1 modulo 3. You can do the same for $p=5$
A: $$2x^2-1=2(x^2-1)+1$$ Therefore by: $$2m+1\mid 2n+1\implies n\equiv m\pmod {2m+1}$$ We get :$$x^2-1\equiv m\pmod{2m+1}\implies x^2\equiv m+1\pmod {2m+1}$$  Since $3=2(1)+1$ we would need $2$ to be a quadratic residue modulo 3. Likewise, $5=2(2)+1$ would require $3$ to be a quadratic residue modulo 5. Since neither is true, we get neither can divide the expression. Two is already ruled out because it's always odd.
For Mersenne numbers greater than 3, we have they can only be prime if they divide a specific member of the iterated sequence starting with $x=2$ this is mostly because $(2x)^2-2\equiv 0\pmod q\implies 2x^2-1\equiv 0 \pmod q$ as we can multiply by the multiplicative inverse of 2 modulo $q$ any time $q$ is odd.
